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I am new to DSP, and I'm self studying. I am confused about the magnitude of $e^{j\omega}$ - where $\omega$ is the normalized angular frequency - when we are on the unit circle.

According to the text book, it's $|e^{j\omega}|=1$. But why is that? I'd really appreciate it is someone could please kindly explain this to me.

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  • $\begingroup$ My first blog article titled "The Exponential Nature of the Complex Unit Circle" starts with the definition of $i$ (aka $j$) and uses simple algebra to work up to an understanding of Euler's equation which is the answer you are looking for. You can find my article here: dsprelated.com/showarticle/754.php $\endgroup$ Nov 23 '18 at 2:30
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Warning: $|e^{j\omega}|$ is equal to $1$ if and only if $\omega$ is a real number. More generally, for $z$ complex, $|e^{z}|=e^{\Re{z}}$.

This really depends in your prior knowledge, because functions like exponentials, sines or cosines, can be developed in different ways. Here, I'll assume you know about sines and cosines, as you refer to angular frequency. Then, an historical way to define the exponential is via Euler formula, for instance:

$$e^{j\omega} = \cos \omega+j\sin \omega\,.$$

Then, through Argand's complex plane interpretation, you can interpret a complex number as a point with 2D coordinates $(x,y) \leftrightarrow x+jy$: the real part refers to the $X$-axis, the imaginary one to the $Y$-axis. Thus, obviously, $e^{j\omega}$ has the coordinates of a point on the $(0,0)$-center circle with radius $1$, hence the magnitude $|e^{j\omega}|$ is equal to the radius, or $(\cos \omega)^2+(\sin \omega)^2=1^2$.

In more modern versions, $e^z$ is defined by a series (see Proof of complex conjugate symmetry property of DFT), and $\cos$ and $\sin$ are derived for it, so they don't come first.

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    $\begingroup$ “In more modern versions”, I like that! $\endgroup$
    – Gilles
    Nov 22 '18 at 9:27
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By Euler's formula, assuming $\omega$ is a real number: $$ e^{j \omega} = \cos(\omega) + j \sin(\omega) $$

The definition of magnitude for a complex number $z = x + jy$ is: $$ |z| = \sqrt{x^2 + y^2}, $$ therefore: $$ \left|e^{j \omega} \right| = \sqrt{\cos^2\omega + \sin^2\omega} = 1 $$ by trigonometric identity.

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    $\begingroup$ Always, as long as $\omega$ is a real number $\endgroup$ Nov 21 '18 at 20:54
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    $\begingroup$ @LaurentDuval eventhough you are spot on your warning on the type of variable being used in the exponential, as you can remember at the very beginning of Diff.Eq., Signals&Systems, etc. the quantitty was first introduced as the general complex exponential : $e^s$ with $s = \sigma + j \omega$ with real $\sigma,\omega$ and then $e^{s} = e^{\sigma + j\omega}$ was the natural consequent, upon which the interpretation of $e^{j \omega}$ needed no more re-definition for $\omega$ to be real. hence I think un-defining $\omega$ as complex would be against the Occam's Razor principle. ;-) $\endgroup$
    – Fat32
    Nov 21 '18 at 22:35
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    $\begingroup$ even though it was not explicitly given as an axiom, i had always considered $\Im\{\omega\}=0$ as an implicit axiom. $\endgroup$ Nov 22 '18 at 7:31
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    $\begingroup$ @LaurentDuval, i still don't understand where the presumption that $\omega$ might not be real in $e^{j\omega}$ comes from. $\endgroup$ Nov 23 '18 at 9:00
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    $\begingroup$ I understand the need to be complete, but I think it's clear from context that $\omega$ is angular frequency (real quantity). If it was $\sigma$, or some other variable, I would agree that more explanation would be needed. I'll go ahead and edit my answer to put this to bed $\endgroup$
    – Robert L.
    Nov 23 '18 at 18:31
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$Ke^{j\theta}$ is a mathematical way of describing a phasor with magnitude K and angle $\theta$. I think knowing this simple relationship takes a lot of the mystery away; which is proven by Euler's identity as Carlos has shown.

What is interesting is if you raise ANY real number to the power of j the magnitude will be 1!

You can see this by solving for the magnitude of a complex number which is the square root of the complex conjugate multiplication:

$\sqrt{x^j x^{-j}} = \sqrt{x^0} = 1$

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